Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,

Consider the following matrix:

[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.

Given target = 20, return false.

解：

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        int m,n;        m=matrix.size();        if(m==0)return false;        n=matrix[0].size();        int i=0;        int l=0,r=n-1;        while(i<m&&l<=r)        {            vector<int>& arr=matrix[i];            //这里注意用二分搜索一个lower bound的时候对要搜索的区间进行单向收缩，因此需要注意1. 只剩两个元素时为了使区间正确收缩到只剩一个元素，需要用对中点mid取上界（ceil函数）2. 只剩一个元素的时候退出循环，因此要额外检测最初给出的数列是不是只包含一个元素，进行元素大小的检测            while(l<r)            {                int mid_index=ceil((l+r)/2.);                int mid=arr[mid_index];                if(mid==target)return true;                else if(mid>target)r=mid_index-1;                else l=mid_index;            }            if(arr[l]>target)return false;            if(arr[l]==target)return true;            r=m-1;            if(matrix[r][l]<target)return false;            if(matrix[r][l]==target)return true;            while(i<r)            {                int mid=matrix[(i+r)/2][l];                int mid_index=(i+r)/2;                if(mid==target)return true;                else if(mid<target)i=mid_index+1;                else r=mid_index;            }            r=l;            l=0;        }        return false;    }};